Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Review - Exercises - Page 390: 105

Answer

$7.886$

Work Step by Step

Using relation $pH=-log[H^{+}]$ This implies that $pH=-log[1.3 \times 10^{8}] \implies -\log 1.3 -\log 10^{-8}$ Thus, $pH=8 \log (10)-0.114=8(1)-0.114=7.886$
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